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How to prove Galilean invariance

  1. May 31, 2012 #1

    Wox

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    I'm having problems showing that Newton's second law of motion stays invariant (has the same form) under a Galilean transformation. If we write the general Galilean transformation as
    [itex]t=t'+t_{t}[/itex]
    [itex]\bar{x}=R\bar{x}'+\bar{u}t'+\bar{t}_{\bar{x}}[/itex]
    where R an orthogonal transformation, then velocity and acceleration transform like this

    [itex]\bar{v}(t)=\frac{d\bar{x}}{dt'}\frac{dt'}{dt} =R\bar{v}'(t')+\bar{u}[/itex]
    [itex]\bar{a}(t)=\frac{d\bar{v}}{dt'}\frac{dt'}{dt} =R\bar{a}'(t')[/itex]

    Therefore we can write for Newton's second law of motion
    [itex]\bar{F}=m\bar{a}(t)=mR\bar{a}'(t')[/itex]

    If this laws is invariant under a Galilean transformation then
    [itex]\bar{F}'=m\bar{a}'(t')[/itex]

    In other words, proving that Newton's second law of motion is invariant under a Galilean transformation, reduces to proving that
    [itex]\bar{F}=R\bar{F}'[/itex]

    However, I do not see why this should be the case. Can anyone shed some light on this?
     
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  3. May 31, 2012 #2

    vanhees71

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    Your proof is correct. All vectors/vector fields transform under rotations as indicated, and the force is a vector.
     
  4. May 31, 2012 #3

    Wox

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    I don't get it, a vector transforms like [itex]\bar{x}=R\bar{x}'+\bar{u}t'+\bar{t}_{\bar{x}}[/itex]
     
  5. May 31, 2012 #4
    Don't know what that equation means, but to prove that the Second law of motion is invariant under a Galilean transformation,see this.
     
  6. May 31, 2012 #5

    Wox

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    The equation is the general Galilean transformation (at least the spatial part, see original post). The source you refer too is exactly the way this is normally shown in textbooks (with [itex]R=id, t_{t}=0[/itex] and [itex]\bar{t}_{\bar{x}}=\bar{0}[/itex]), but then I still don't understand it. They start by assuming that Newton's second law of motion holds in the primed reference frame [itex]f'=ma'[/itex]. Ok, so we need to show that it also holds in the unprimed frame (which is related by a Galilean transformation to the primed frame). So we need to show that [itex]f=ma[/itex] holds, knowing the [itex]f'=ma'[/itex] holds and that the frames are related by a Galilean transformation.

    I follow the steps until [itex]f'=ma'=....=ma[/itex] (now without [itex]R[/itex] as in the original post, because we assumed it to be the identity). But then the next step [itex]f'=ma'=....=ma=f[/itex] I don't understand. We don't know yet whether [itex]f=ma[/itex] holds in the unprimed frame. In fact, this is exactly what we're trying to prove.
     
    Last edited: May 31, 2012
  7. May 31, 2012 #6
    Is R a scaling factor?
    You're kinda right there. Now I'm confused too.
     
  8. May 31, 2012 #7

    Wox

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    R is an orthogonal transformation in Euclidean vector space, therefore it is a combination of rotations.
     
  9. May 31, 2012 #8
    I think you should now pick a specific force law, such [tex]\vec{F} = \frac{GMm}{|\vec{x_1} - \vec{x_2}|^3}(\vec{x_1} - \vec{x_2})[/tex] for Newtonian gravity. Working from the explicit equation for the force you can confirm that it obeys this transformation law for F. It's perfectly possible to write down force laws, for example a drag force [tex]\vec{F} = -\frac{1}{2}c v^2 \hat{v}[/tex], which do not satisfy the transformation law you want. Your reasoning then shows that such force laws cannot lead to Galilean invariant physics. For example, the drag force above only takes that simple form in the special frame of reference at rest with respect to the medium producing the drag
     
  10. Jun 1, 2012 #9

    Wox

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    I have read several sources and they clearly state that "Newton's Second Law of Motion is invariant under the Galilean Transformation". Therefore one should prove it, regardless the nature of the force.

    Maybe using 4-vectors like in special relativity solves the problem? If we define a world line in Galilean space-time [itex]\mathbb{R}^{4}[/itex] as the following curve
    [itex]\bar{w}\colon I\subset\mathbb{R}\to \mathbb{R}^{4}\colon t\mapsto (t,\bar{x}(t))[/itex]
    [itex]\bar{x}\colon I\subset\mathbb{R}\to \mathbb{R}^{3}\colon t\mapsto (x(t),y(t),z(t))[/itex]

    where [itex]\mathbb{R}^{3}\subset\mathbb{R}^{4}[/itex] Euclidean, then the acceleration is given by

    [itex]\bar{a}\colon I\subset\mathbb{R}\to \mathbb{R}^{4}\colon t\mapsto \frac{d^{2}\bar{w}(t)}{dt^{2}}=(0,\frac{d^{2}\bar{x}(t)}{dt^{2}})\equiv(0,\tilde{a}(t))[/itex]

    where [itex]\tilde{a}[/itex] the classical acceleration and the force field that causes the acceleration

    [itex]\bar{F}\colon\mathbb{R}^{4}\to\mathbb{R}^{4}\colon\bar{w}(t)\mapsto m\bar{a}(t)=(0,m\tilde{a}(t))[/itex]

    So if Newton's second law of motion is written as [itex]\bar{F}(\bar{w})=m\bar{a}(t)[/itex] then a Galilean transformation causes [itex]\bar{F}=G\bar{F}'[/itex] and [itex]\bar{a}=G\bar{a}'[/itex] since they both live in [itex]\mathbb{R}^{4}[/itex]. Therefore [itex]\bar{F}(\bar{w})=m\bar{a}(t)\Rightarrow G\bar{F}'(\bar{w})=mG\bar{a}'(t)\Rightarrow \bar{F}'(\bar{w})=m\bar{a}'(t)[/itex] which is what we needed to prove.

    However this is not compliant with the reasoning in the original post. A Galilean transformation in [itex]\mathbb{R}^{4}[/itex] given by
    [itex]t=t'+t_{t}[/itex]
    [itex]\bar{x}=R\bar{x}'+\bar{u}t+\bar{t}_{\bar{x}}[/itex]

    [itex]\bar{v}(t)=\frac{d\bar{w}(t)}{dt}=\frac{d\bar{w}(t)}{dt'}\frac{dt'}{dt}=\frac{d\bar{w}(t)}{dt'}= (1,R\frac{d\bar{x}'}{dt'}+\bar{u})[/itex]
    [itex]\bar{a}=(0,\tilde{a}(t))=\frac{d\bar{v}(t)}{dt}=(0,R\frac{d^{2}\bar{x}'}{dt'^{2}})=(0,R\tilde{a}'(t'))[/itex]
    [itex]\tilde{a}(t)=R\tilde{a}'(t')[/itex]

    This is not the same as [itex]\bar{a}=G\bar{a}'[/itex] because

    [itex]\bar{a}(t)=G\bar{a}'(t')[/itex]
    [itex]\Leftrightarrow\begin{pmatrix} 0 \\ \tilde{a}(t) \\ 1\end{pmatrix} = \begin{pmatrix}1&0&t_{t}\\
    \bar{u}&R&t_{\bar{x}}\\
    0&0&1
    \end{pmatrix}\cdot\begin{pmatrix} 0 \\ \tilde{a}'(t') \\ 1\end{pmatrix}[/itex]
    [itex]\Leftrightarrow t_{t}=0[/itex] and [itex] \tilde{a}(t)=R\tilde{a}'(t')+t_{\bar{x}}[/itex]

    So we have an extra translation [itex]t_{\bar{x}}[/itex] which should not be there. Any thoughts?
     
  11. Jun 4, 2012 #10
    I figured it out. The force and mass are taken to be invariant. The acceleration as seen from both frames is the same, a'=a. Now, F=ma holds in one frame. Since F' and m' are same as F and m resp, F'=m'a'
     
  12. Jun 6, 2012 #11

    Wox

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    As shown in the original post, a is not invariant as [itex]\bar{a}(t)=R\bar{a}'(t')[/itex].

    Maybe the following is a valid reasoning. The relation between the coordinates [itex](t,x,y,z)[/itex] of an event in Galilean space-time (which is an affine space) with respect to two reference frames which are related by a Galilean transformation, is given by

    [itex]t=t'+t_{t}[/itex]
    [itex]\bar{x}=R\bar{x}'+\bar{u}t'+\bar{t}_{\bar{x}}[/itex]

    Now the classical acceleration is defined as [itex]\bar{a}(t)=\frac{d\bar{x}}{dt}[/itex] where [itex]\bar{x}(t)[/itex] doesn't live in Galilean space-time. First of all it lives in the associated vector space of Galilean space-time, therefore the origin shift [itex](t_{t},\bar{t}_{\bar{x}})[/itex] doesn't have an effect on it. Secondly, it isn't just a vector in Galilean vector space (which is 4 dimensional) but it is a vector in a 3D subspace (which is Euclidean, as opposed the Galilean vector space itself). Therefore the boost [itex]\bar{u}[/itex] has no effect on it (since it involves the 4th dimension). What is left is [itex]R[/itex] and in fact as shown in the original post, a Galilean change of frame relates the acceleration as observed in both frames as

    [itex]\bar{a}(t)=R\bar{a}'(t')[/itex]

    Suppose that we define a vector field

    [itex]\bar{F}\colon\mathbb{R}^{4}\to\mathbb{R}^{3}\colon \bar{w}(t)\mapsto m\bar{a}(t)[/itex]

    then [itex]\bar{F}[/itex] lives in the same space as [itex]\bar{a}[/itex] and therefore transforms like this

    [itex]\bar{F}(\bar{w}(t))=R\bar{F}'(\bar{w}'(t'))[/itex]

    Since we already know that [itex]\bar{F}(t)=m\bar{a}(t)=mR\bar{a}'(t')[/itex] it follows that

    [itex]\bar{F}'(\bar{w}'(t'))=m\bar{a}'(t')[/itex]

    Therefore it doesn't matter in which frame you defined the vector field [itex]\bar{F}[/itex], it will have the same form in all other inertial frames. So if force would be defined as such a vector field, one could say that it's definition is Galilean invariant. However force is not defined like this, it only obeys this law. In that case you should check whether F transforms with [itex]R[/itex] and when it does, you can say that the relation F=ma is invariant under a Galilean transformation, but ONLY for this particular force and not in general.
     
  13. Jun 18, 2012 #12
    It looks to me that this is where you created an big and useless complication. The Galilean transformation has no rotations - there is just a translation along X.
     
  14. Jun 18, 2012 #13

    vanhees71

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    The full Galilei group consists indeed of rotations, boosts and space-time translations. To be more precise, it's a semidirect product of the corresponding subgroups. It's a 10 dimensional Lie group, leading to the 10 conserved quantities, angular momentum (3 components), center-of-mass velocity (3 components), momentum (3 components), and energy (1 component).
     
  15. Jun 18, 2012 #14
    Probably there is no limit to how complex one can make it. Commonly "Galilean invariance" simply refers to the "Galilean transformation"
    http://en.wikipedia.org/wiki/Galilean_invariance

    From there on, is there any reason to doubt that one can next also rotate etc?
     
  16. Jun 19, 2012 #15

    Wox

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    I think considering the orthogonal part R (rotations) of the Galilean transformation is important here. Without it, the acceleration measured in two inertial frames is the same (because a'=R.a). Therefore one often jumps to the conclusion that F=m.a is Galilean invariant. But the logic escapes me. These are the three things we know

    unprimed inertial frame:
    1. [itex]\bar{F}=m\bar{a}[/itex] holds

    primed inertial frame:
    2. relates to the unprimed frame by
    [itex]\quad\quad t=t'+t_{t}[/itex]
    [itex]\quad\quad\bar{x}=R\bar{x}'+\bar{u}t'+ \bar{t}_{\bar{x}}[/itex]
    3. [itex]\bar{a}'=R\bar{a}[/itex] (follows from 2. by [itex]\bar{a}=\frac{\bar{d^{2}\bar{x}}}{dt^{2}}[/itex] and [itex]\bar{a}'=\frac{\bar{d^{2}\bar{x}'}}{dt'^{2}}[/itex])

    Now we need to prove that [itex]\bar{F}'=m\bar{a}'[/itex] holds. Or in other words that [itex]\bar{F}'=mR\bar{a}[/itex]. Now we can prove that for a specific force, but since one says the Newton's second law of motion is invariant under a Galilean transformation (i.e. has the same form in all inertial frames), we have to prove it regardless the nature of the force. I'm starting to think that the phrase "Newton's second law of motion is Galilean invariant" doesn't make any sense. Maybe one can only say that a particular force is Galilean invariant or not (by checking that [itex]\bar{F}'=mR\bar{a}[/itex])? Does anyone know of an article/book where this subject is treated rigorously? I often see it just mentioned as if it was trivial.
     
  17. Jun 19, 2012 #16
    I have a problem with either your notation or your starting claims. Using the notation that I'm familiar with (Alonso&Finn, etc.):

    1. [itex]\bar{F}=m\bar{a}[/itex]

    2. [itex]\bar{r}=\bar{r} - \bar{v} t[/itex]

    In the "Galilean" transformation we choose v along X:

    3. x'=x-vt, y'=y, z'=z, t'=t

    The assumptions that allows to do this, are such things as homogeneity and isotropy of space - the same assumptions on which (1) is based.

    Thus, similar to vin300, I don't find your R. You seem to ask if we can prove that if space is homogeneous and isotropic before a linear reference system transformation, that it is also like that after the transformation...
     
  18. Jun 20, 2012 #17

    Wox

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    A Galilean transformation is an automorphism of Galilean space-time [itex]\mathbb{R}^{4}[/itex] (which is an affine space). In other words, a Galilean transformation is an invertible map [itex]G\colon \mathbb{R}^{4}\to\mathbb{R}^{4}[/itex] which "preserves the structure" of Galilean space-time. But what structure?

    Compare this with what we are most familiar with: Euclidean space. Suppose we consider [itex]\mathbb{R}^{3}[/itex] to be Euclidean space (which is an affine space). Then the "structure" to be preserved is the Euclidean structure, or more precisely, distances and angles between points. Therefore the automorphisms of Euclidean space, known as isometries, can be written (in homogeneous coordinates) as
    [tex]
    \begin{bmatrix}
    x\\
    y\\
    z\\
    1\\
    \end{bmatrix}=\begin{bmatrix}R_{11}&R_{12}&R_{13}&t_{x}\\
    R_{21}&R_{22}&R_{23}&t_{y}\\
    R_{31}&R_{32}&R_{33}&t_{z}\\
    0&0&0&1
    \end{bmatrix}\begin{bmatrix}
    x'\\
    y'\\
    z'\\
    1\\
    \end{bmatrix}
    [/tex]
    or in short
    [tex]
    \begin{bmatrix}
    \bar{x}\\
    1\\
    \end{bmatrix}=\begin{bmatrix}R&\bar{t}_{\bar{x}}\\
    0&1
    \end{bmatrix}\begin{bmatrix}
    \bar{x}'\\
    1\\
    \end{bmatrix}
    [/tex]
    where [itex]R[/itex] an orthogonal transformation (e.g. rotation). By we don't have 3D Euclidean space, we have 4D Galilean space-time. The associated vector space of Galilean space-time has 3D Euclidean vector space as subspace and the complement is the time dimension. The structure that needs to be preserved is the Euclidean one in a 3D subspace and time difference between events. Therefore automorphisms of Galilean space-time, known as Galilean transformations, can be written as (for convenience we choose the basis so that the last three basis vectors span the 3D Euclidean subspace)
    [tex]
    \begin{bmatrix}
    t\\
    x\\
    y\\
    z\\
    1\\
    \end{bmatrix}=\begin{bmatrix}
    1&0&0&0&t_{t}\\
    u_{x}&R_{11}&R_{12}&R_{13}&t_{x}\\
    u_{y}&R_{21}&R_{22}&R_{23}&t_{y}\\
    u_{z}&R_{31}&R_{32}&R_{33}&t_{z}\\
    0&0&0&0&1
    \end{bmatrix}\begin{bmatrix}
    t\\
    x'\\
    y'\\
    z'\\
    1\\
    \end{bmatrix}
    [/tex]
    or in short
    [tex]
    \begin{bmatrix}
    t\\
    \bar{x}\\
    1\\
    \end{bmatrix}=\begin{bmatrix}
    1&0&t_{t}\\
    \bar{u}&R&\bar{t}_{\bar{x}}\\
    0&0&1
    \end{bmatrix}\begin{bmatrix}
    t'\\
    \bar{x}'\\
    1\\
    \end{bmatrix}
    [/tex]
    for more details: see here. I prefer these kind of rigorous treatments because phrases as "homogeneity and isotropy of space" always seem very vague to me.

    On the side: I find this very useful to understand why we state that all physical laws should be invariant under a Galilean transformation. A Galilean transformation preserves the structure of Galilean space-time, therefore all laws that are defined in this space-time, must also be preserved after a Galilean transformation (= change of reference frame). From this we can also see why we only consider observers in mutual uniform motion ([itex]\bar{u}[/itex]) and claim that all laws should have the same form for them. In other words, the principle of relativity (Galilean in this case) is inherent to the structure of space-time and not an additional postulate.
     
    Last edited: Jun 20, 2012
  19. Jun 21, 2012 #18
    R will be a unitary matrix (all rotation matrices are). By definition a unitary matrix will preserve length (or magnitude) of your vector. So your proof is basically saying that if you are rotating your reference frame, your force is now going to be pointing in a new direction. That sounds right to me :).
     
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